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Particle shot from a height h on Earth

  1. Sep 17, 2016 #1
    1. The problem statement, all variables and given/known data
    A cannon that is capable of firing a shell at speed [tex]V_0[/tex] is mounted on a vertical tower of height h that overlooks a level plain below.

    (a) Show that the elevation angle [tex]\alpha[/tex] at which the cannon must be set to achieve maximum range is given by the expression

    [tex]csc^2(\alpha)=2(1+\frac {gh} {V_0^2})[/tex]

    (b) What is the maximum range R of the cannon?


    2. Relevant equations

    [tex]F=m\frac {d^2r} {dt^2}[/tex]

    3. The attempt at a solution

    The only force on the particle is -mgk, so

    [tex]m\frac{d^2r} {dt^2} = -mgk[/tex]

    [tex]\frac{d^2r} {dt^2} = -gk[/tex]

    [tex]\vec{v}=-gtk+v_0[/tex]
    [tex]\vec{r}=\frac {-gt} {2}k+v_0t+r_0[/tex]

    The initial velocity and position can be expressed as such

    [tex]V_0=V_0cos(\alpha)i+V_0sin(\alpha)k[/tex]
    [tex]r_0=hk[/tex]

    Therefore vector r is

    [tex]\vec{r}=V_0tcos(\alpha)i+(h+V_0tsin(\alpha)-\frac{gt^2} {2})k[/tex]

    In component form,

    [tex]x=V_0tcos(\alpha)[/tex]
    [tex]z=(h+V_0tsin(\alpha)-\frac {gt^2} {2})k[/tex]

    when z=0 (taking just the postitive root of the quadratic formula because we are concerned about positive t)

    [tex]t= \frac {V_0sin(\alpha) + \sqrt{V_0^2sin^2(\alpha)+2gh}} {g}[/tex]

    Plugging this into x, we get

    [tex]x=V_0cos(\alpha)(\frac {V_0sin(\alpha)+\sqrt{V_0^2sin^2(\alpha)+2gh}} {g})[/tex]

    This is the equation of motion in the x direction. It seems to make sense because if the angle is 90 degrees, we get 0 distance traveled. When the angle is 0 degrees, the y component of the initial velocity cancels out to give something nice. My thinking from this point is that taking dx/da and setting it equal to 0 would give the angle that would make x a maximum. However, when I do this I get a horribly complicated equation that I can't figure out how to simplify. Am I on the right track?
     
  2. jcsd
  3. Sep 17, 2016 #2

    mfb

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    Check the second equation. You got it right later so I guess it is just a typo.

    The last equation looks good. The value of ##\alpha## maximizing this also corresponds to a value of ##\cos(\alpha)## maximizing this (and due to the range of ##\alpha## you don't have to deal with ambiguities like ##\pm 2 \pi##), so you can introduce a new variable ##p=\cos(\alpha)## and maximize x with respect to this one. That should make the expressions easier. Choosing ##p=\sin(\alpha)## could work as well.

    You know the right answer already. You do not have to solve the equation, you just have to show that the answer works.
     
  4. Sep 18, 2016 #3
    So

    [tex]\frac {dp} {d\alpha} = -sin\alpha[/tex]

    That would be correct right? I don't think I've done something like this exactly before. So if I plugged it in and got something like

    [tex]\frac {dp} {d\alpha}p[/tex]

    That would equal one right? Sorry if that's a dumb question.
     
  5. Sep 18, 2016 #4

    mfb

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    Yes, but it does not matter. If you substitute before taking the derivative with respect to p ##\alpha## does not appear any more.

    -sin(α)cos(α) is not equal to 1.
     
  6. Sep 19, 2016 #5

    SammyS

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    Assuming the last equation is correct, the following may help.

    Solve for ##\displaystyle \ \sqrt{V_0^2\sin^2(\alpha)+2gh} \,.\ ##

    Then square both sides and do a little cancelling/simplifying.

    Use implicit differentiation & set ##\ dx / d\alpha =0 \ . ##

    Added in Edit:

    Thanks mfb (for the "like"), but it didn't work out as well as I had hoped. Maybe I just didn't find a good way to eliminate x in the result .

    You can get essentially the same thing my suggestion leads to, by completing the square in the expression for time, z = 0, then substituting the expression for x as a function of t. (Probably should be labeled t0 for the time to reach zero elevation.)
     
    Last edited: Sep 19, 2016
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